Tricky Fourier Transform problem for an exponential function

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Homework Statement


find the fourier transform, using the definition of the fourier transform [itex]\widehat{f}[/itex]([itex]\nu[/itex])=∫[itex]^{∞}_{-∞}[/itex]f(t)e[itex]^{-2 \pi i \nu t}[/itex]dt, of the function f(t)=2 [itex]\pi[/itex]t[itex]^{2}[/itex]e[itex]^{- \pi t^{2}}[/itex]


Homework Equations



I have the answer:

(1-2[itex]{\pi \nu^{2}}[/itex])e[itex]^{- \pi \nu^{2}}[/itex]

The Attempt at a Solution



After inserting f(t) into the equation for the transform, I added the exponents on the e terms, factored out -[itex]\pi[/itex], and added and subtracted [itex]\nu^{2}[/itex] to get

∫[itex]^{∞}_{-∞}[/itex]2[itex]\pi[/itex]t[itex]^{2}[/itex]e[itex]^{- \pi (t^{2}+2 i \nu t + \nu^{2} - \nu^{2})}[/itex]dt

I then substituted x=t+i[itex]\nu[/itex] to get

∫[itex]^{∞}_{-∞}[/itex]2[itex]\pi[/itex](t+i[itex]\nu[/itex])[itex]^{2}[/itex]e[itex]^{- \pi (x^{2}+ \nu^{2} )}[/itex]dt

at this point I know that I should be able to solve the integral, probably by integrating by parts, but I am really just lost. This seems like such a tricky integral. I thought maybe squaring the equation to get a double integral and then converting to polar coordinates would work but I couldn't get that to work out either. Thanks for the help!
 
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Answers and Replies

  • #2
Ray Vickson
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Homework Statement


find the fourier transform, using the definition of the fourier transform [itex]\widehat{f}[/itex]([itex]\nu[/itex])=∫[itex]^{∞}_{-∞}[/itex]f(t)e[itex]^{-2 \pi i \nu t}[/itex]dt, of the function f(t)=2 [itex]\pi[/itex]t[itex]^{2}[/itex]e[itex]^{- \pi t^{2}}[/itex]


Homework Equations



I have the answer:

(1-2[itex]{\pi \nu^{2}}[/itex])e[itex]^{- \pi \nu^{2}}[/itex]

The Attempt at a Solution



After inserting f(t) into the equation for the transform, I added the exponents on the e terms, factored out -[itex]\pi[/itex], and added and subtracted [itex]\nu^{2}[/itex] to get

∫[itex]^{∞}_{-∞}[/itex]2[itex]\pi[/itex]t[itex]^{2}[/itex]e[itex]^{- \pi (t^{2}+2 i \nu t + \nu^{2} - \nu^{2})}[/itex]dt

I then substituted x=t+i[itex]\nu[/itex] to get

∫[itex]^{∞}_{-∞}[/itex]2[itex]\pi[/itex](t+i[itex]\nu[/itex])[itex]^{2}[/itex]e[itex]^{- \pi (x^{2}+ \nu^{2} )}[/itex]dt

at this point I know that I should be able to solve the integral, probably by integrating by parts, but I am really just lost. This seems like such a tricky integral. I thought maybe squaring the equation to get a double integral and then converting to polar coordinates would work but I couldn't get that to work out either. Thanks for the help!
Your last expression is wrong: it should be [itex] \int_{-\infty + i \nu}^{\infty + i \nu} 2 \pi x^2 e^{-\pi (x^2 + \nu^2)}\, dx , [/itex], which can be replaced by the same integral from x = -infinity to +infinity along the real x-axis (why)? Now int x^2*exp(-x^2) dx can be attacked using integration by parts.

RGV
 
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